Problem: The side of a cube is increasing at a rate of $2$ kilometers per hour. At a certain instant, the side is $1.5$ kilometers. What is the rate of change of the volume of the cube at that instant (in cubic kilometers per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $8$ (Choice C) C $13.5$ (Choice D) D $16.5$
Setting up the math Let... $s(t)$ denote the cube's side at time $t$, and $V(t)$ denote the cube's volume at time $t$. We are given that $s'(t)=2$ and that $s(t_0)=1.5$ for a specific time $t_0$. We want to find $V'(t_0)$. Relating the measures $V(t)$ and $s(t)$ relate to each other through the formula for the volume of a cube: $V(t)=[s(t)]^3$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=3[s(t)]^2s'(t)$ Using the information to solve Let's plug ${s(t_0)}={1.5}$ and ${s'(t_0)}={2}$ into the expression for $V'(t_0)$ : $\begin{aligned} V'(t_0)&=3[{s(t_0)}]^2{s'(t_0)} \\\\ &=3({1.5})^2({2}) \\\\ &=13.5 \end{aligned}$ In conclusion, the rate of change of the volume of the cube at that instant is $13.5$ cubic kilometers per hour. Since the rate of change is positive, we know that the volume is increasing.